Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Chemistry guide (first assessment 2025).
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S3.2.1 Empirical formula
Empirical formula
The simplest whole-number ratio of the atoms of each element in a compound. Butane, C₄H₁₀, has the empirical formula C₂H₅; ethanoic acid, C₂H₄O₂, has the empirical formula CH₂O. When the numbers in a molecular formula have no common factor, as in C₃H₈O, the empirical and molecular formulas are the same.
Molecular formula
The actual number of atoms of each element in one molecule of a compound, a whole-number multiple of the empirical formula (for example C₆H₁₂O₆ for glucose, empirical formula CH₂O). A molecular formula does not show how the atoms are connected, so several compounds (isomers) can share one molecular formula: C₃H₈O is propan-1-ol, propan-2-ol and methoxyethane.
Full structural formula
A two-dimensional representation that shows every atom in the molecule and every bond between atoms, each covalent bond drawn as a line (a double bond as two lines). It shows connectivity but not the true three-dimensional shape.
Condensed structural formula
A formula that shows, carbon by carbon, the atoms or groups attached to each carbon atom without drawing all the bonds, for example CH₃CH₂CH₂OH for propan-1-ol and CH₃CH(OH)CH₃ for propan-2-ol. Brackets show a branch or group attached to the preceding carbon atom. It must be unambiguous: it specifies one compound.
Skeletal formula
A simplified structural formula in which the carbon chain is drawn as a zig-zag of lines. Every line end and every vertex (bend) represents a carbon atom, and the hydrogen atoms bonded to carbon are not shown: each carbon is assumed to carry enough hydrogen atoms to make four bonds in total. Atoms other than carbon and hydrogen, and hydrogen atoms bonded to them (as in –OH), are shown explicitly.
Stereochemical formula
A formula that shows the three-dimensional arrangement of atoms. Bonds in the plane of the paper are drawn as plain lines, a bond pointing towards the viewer as a solid tapered wedge, and a bond pointing away from the viewer as a dashed (hashed) wedge; around a tetrahedral carbon atom, two bonds are usually drawn in the plane, one wedged and one dashed.
Students often think An empirical formula is always a smaller, simplified version of the molecular formula, so the two can never be the same. In fact Yes. When the numbers of atoms in the molecular formula have no common factor, as in C₃H₈O or CH₄, the empirical formula is the molecular formula.
Students often think A formula that lists every atom in the molecule is a structural formula, because it shows everything the molecule contains. In fact No. C₃H₈O is a molecular formula: it shows the number of each type of atom, not which atoms are bonded to which. A structural formula such as CH₃CH(OH)CH₃ shows the connectivity.
S3.2.2 Functional group
Functional group
An atom or group of atoms in an organic molecule that gives the compound its characteristic chemical and physical properties. Organic compounds are divided into classes, and homologous series, according to the functional groups their molecules contain.
Halogeno group
A halogen atom (F, Cl, Br or I), written –X, covalently bonded to a carbon atom, as in chloroethane, CH₃CH₂Cl. It is the functional group of the halogenoalkanes.
Hydroxyl group
An –OH group covalently bonded to a carbon atom, the functional group of the alcohols (for example CH₃CH₂OH). It is not the hydroxide ion, OH⁻: the O–H group in an alcohol does not ionize in water to give an alkaline solution.
Carbonyl group
A carbon atom double-bonded to an oxygen atom, C=O. In aldehydes the carbonyl carbon is bonded to at least one hydrogen atom (written –CHO, at the end of a chain); in ketones it is bonded to two carbon atoms (written –CO– within a chain, as in CH₃COCH₃).
Carboxyl group
The group –COOH, in which one carbon atom is double-bonded to an oxygen atom and single-bonded to an –OH group. It is the functional group of the carboxylic acids, for example CH₃COOH. It is treated as a single functional group, not as a separate carbonyl and hydroxyl group.
Alkoxy group
An oxygen atom single-bonded to an alkyl group, –O–R, and so bonded between two carbon atoms in the molecule, as in CH₃CH₂OCH₃ (methoxyethane). It is the functional group of the ethers. The oxygen carries no hydrogen atom.
Amino group
A nitrogen atom bonded to a carbon atom by a single bond and to hydrogen atoms or further alkyl groups, –NH₂, –NHR or –NR₂, with no C=O on the carbon bonded to nitrogen. It is the functional group of the amines, for example CH₃CH₂NH₂.
Amido group
A carbonyl carbon atom bonded directly to a nitrogen atom, –CONH₂ (or –CONH– and –CON< when the nitrogen carries alkyl groups). It is the functional group of the amides, for example CH₃CONH₂ (ethanamide).
Ester group
The group –COO– in which a carbonyl carbon is single-bonded to an oxygen atom that is itself bonded to a carbon atom, R–COO–R′, as in CH₃COOCH₂CH₃ (ethyl ethanoate). It is the functional group of the esters.
Phenyl group
The group –C₆H₅: a benzene ring bonded to the rest of the molecule through one of its carbon atoms, as in methylbenzene, C₆H₅CH₃.
Saturated compound
An organic compound whose molecules contain only single bonds between carbon atoms, for example the alkanes and cycloalkanes (cyclohexane, C₆H₁₂, is saturated).
Unsaturated compound
An organic compound whose molecules contain at least one carbon–carbon double bond (C=C) or triple bond (C≡C), for example the alkenes and alkynes. Each carbon atom in a C=C bond still forms four bonds in total.
Students often think Any compound whose formula contains 'COO' has an ester group, so carboxylic acids and esters are the same class. In fact No. CH₃COOH contains the carboxyl group, –COOH: the singly bonded oxygen carries a hydrogen atom. In an ester, R–COO–R′, that oxygen is bonded to a carbon atom.
Students often think Esters and ethers are the same kind of compound, because both have an oxygen atom joining two carbon chains. In fact An ester contains –COO–, a carbonyl carbon bonded to an oxygen that is bonded to another carbon (R–COO–R′). An ether contains only an oxygen single-bonded between two carbon atoms, R–O–R′, with no C=O.
S3.2.3 Homologous series
Homologous series
A family of organic compounds with the same functional group and the same general formula, in which successive members differ by a common structural unit, typically CH₂. Members have similar chemical properties and show a gradual trend in physical properties.
General formula
A formula in terms of n (the number of carbon atoms) that fits every member of a homologous series. Examples for members with one functional group and no rings: alkanes CₙH₂ₙ₊₂; alkenes CₙH₂ₙ; alkynes CₙH₂ₙ₋₂; halogenoalkanes CₙH₂ₙ₊₁X; alcohols CₙH₂ₙ₊₁OH; aldehydes and ketones CₙH₂ₙO; carboxylic acids and esters CₙH₂ₙO₂. Different series can share a general formula (alcohols and ethers are both CₙH₂ₙ₊₂O), so a general formula alone does not identify a series.
Students often think Any two compounds whose molecular formulas differ by CH₂ are successive members of the same homologous series. In fact No. Ethanal is an aldehyde and propanone is a ketone: they have different functional groups, so they belong to different series even though their formulas differ by CH₂.
Students often think Homologues and isomers are the same idea: compounds with the same functional group are isomers, and isomers are members of one series. In fact No. Successive members of a homologous series differ by CH₂ and so have different molecular formulas. Isomers have the same molecular formula.
S3.2.4 Trend in boiling and melting points in a homologous series
Trend in boiling and melting points in a homologous series
Boiling points increase as the carbon chain lengthens. Each additional CH₂ adds electrons and increases the surface area over which neighbouring molecules are in contact, so the London (dispersion) forces between molecules become stronger and more energy is needed to separate the molecules. No covalent bonds are broken on boiling. The functional group contributes the same type of additional intermolecular force (for example one hydrogen-bonding –OH per alcohol molecule) throughout the series. Melting points also increase overall along a series, though less regularly, because they also depend on how the molecules pack in the solid.
Students often think Every hydrogen atom in a molecule can form a hydrogen bond, so molecules with more hydrogen atoms form more hydrogen bonds. In fact No. Hydrogen bonding in alcohols involves the H of the –OH group. Each molecule of methanol, ethanol, propan-1-ol and butan-1-ol has one –OH group, so hydrogen bonding per molecule is similar along the series.
Students often think Boiling breaks covalent bonds within the molecules, so compounds with more covalent bonds have higher boiling points. In fact The intermolecular forces between molecules (London forces and, for alcohols, hydrogen bonds). The covalent bonds inside the molecules are not broken.
S3.2.5 IUPAC nomenclature
IUPAC nomenclature
The systematic rules of the International Union of Pure and Applied Chemistry for naming compounds. For the compounds named at this level, the name is built from: the stem for the number of carbon atoms in the parent chain (meth-, eth-, prop-, but-, pent-, hex-); the suffix for the functional group (-ol, -al, -one, -oic acid) or, for halogeno groups, a prefix (chloro- etc.); -en- with a locant for a C=C bond; and alkyl branch prefixes (methyl-, ethyl-) with locants.
Parent chain
The longest continuous chain of carbon atoms that contains the carbon of the functional group (and, in the compounds named at this level, the C=C bond). Branches not in this chain are named as alkyl prefixes. The carbon of a –CHO or –COOH group is part of the parent chain and is carbon 1.
Locant
A number showing the position of a functional group, a C=C bond or a branch on the parent chain. The chain is numbered from the end that gives the lowest locant first to the group named as the suffix, then to a C=C bond, then to the prefixes. A C=C bond takes the lower of the two numbers of its carbon atoms. Prefixes are listed in alphabetical order (chloro before methyl), regardless of their locants.
Students often think Carbon atoms are numbered from the left-hand end of the formula as written, like reading a sentence. In fact No. The chain is numbered from whichever end gives the lowest locant to the functional group named as the suffix (or, for halogenoalkanes, the lowest set of locants to the prefixes). The direction in which the formula is written is irrelevant.
Students often think -ol and -al are two spellings of the same ending for oxygen-containing compounds, and either can be used. In fact -ol means a hydroxyl group (alcohol, –OH); -al means an aldehyde group (–CHO, a carbonyl at the end of the chain).
S3.2.6 Structural isomers
Structural isomers
Compounds with the same molecular formula but different connectivities (different sequences of bonded atoms). Two drawings or formulas that can be turned into each other by rotating the molecule or rotating about single bonds show the same compound, not isomers.
Chain (branched and straight-chain) isomers
Structural isomers that differ in the carbon skeleton: a straight chain or branched chains. For example butane, CH₃CH₂CH₂CH₃, and 2-methylpropane, CH₃CH(CH₃)CH₃, are both C₄H₁₀.
Position isomers
Structural isomers with the same carbon skeleton and the same functional group but with the group at a different position, for example propan-1-ol, CH₃CH₂CH₂OH, and propan-2-ol, CH₃CH(OH)CH₃.
Functional group isomers
Structural isomers that belong to different homologous series, for example propanal, CH₃CH₂CHO, and propanone, CH₃COCH₃ (both C₃H₆O), or butan-1-ol and ethoxyethane (both C₄H₁₀O).
Primary, secondary and tertiary alcohols and halogenoalkanes
Classified by the number of carbon atoms bonded to the carbon atom that carries the –OH or halogen atom: one (primary, e.g. CH₃CH₂CH₂OH), two (secondary, e.g. CH₃CH(OH)CH₃) or three (tertiary, e.g. (CH₃)₃COH). Methanol and halogenomethanes, with no carbon attached, are usually grouped with the primary compounds.
Primary, secondary and tertiary amines
Classified by the number of carbon atoms bonded directly to the nitrogen atom: one (primary, e.g. CH₃CH₂NH₂), two (secondary, e.g. CH₃NHCH₃) or three (tertiary, e.g. (CH₃)₃N). This differs from the rule for alcohols: (CH₃)₃CNH₂ is a primary amine even though its nitrogen is on a carbon bonded to three other carbons.
Students often think Primary, secondary and tertiary describe the position of the functional group: on carbon 1, carbon 2 or carbon 3 of the chain. In fact No. It is a secondary alcohol: the carbon carrying –OH is bonded to two other carbon atoms. The locant 3 is only a position in the chain.
Students often think An alcohol is tertiary if its molecule contains a carbon atom bonded to three other carbon atoms anywhere in the chain. In fact No. It is a primary alcohol: the carbon bearing –OH (the CH₂) is bonded to only one other carbon. The branched carbon does not carry the –OH.
S3.2.7 Stereoisomers HL
Stereoisomers
Compounds with the same constitution (the same atoms, connectivities and bond multiplicities) but different arrangements of the atoms in space. Cis–trans isomers and enantiomers are stereoisomers.
Cis–trans isomerism
Stereoisomerism caused by restricted rotation, about a C=C bond or within a ring, when each of the two carbon atoms concerned carries two different groups. In the cis isomer, the two reference groups (for example the two CH₃ groups in but-2-ene) are on the same side of the double bond or ring; in the trans isomer they are on opposite sides. But-1-ene and 2-methylpropene do not show it; but-2-ene and 1,2-dimethylcyclopropane do. Free rotation about C–C single bonds in open chains means such chains cannot show it.
Chiral carbon atom
A carbon atom bonded to four different atoms or groups, for example carbon 2 in butan-2-ol, CH₃CH(OH)CH₂CH₃ (bonded to H, OH, CH₃ and CH₂CH₃). A molecule with one chiral carbon exists as two stereoisomers that are non-superimposable mirror images.
Enantiomers
A pair of stereoisomers that are non-superimposable mirror images of each other, shown using wedge-dash formulas with tapered bonds around the chiral carbon. Enantiomers have identical physical properties (melting point, boiling point, density) except for the direction in which they rotate plane-polarized light, and identical chemical properties except in chiral environments, for example with other chiral molecules such as enzymes, where they behave differently.
Optical activity
The ability of a substance to rotate the plane of plane-polarized light. The two enantiomers of a chiral compound rotate the plane by equal angles in opposite directions (under the same conditions).
Racemic mixture
A mixture containing equal amounts of the two enantiomers of a chiral compound. The rotations caused by the two enantiomers cancel, so a racemic mixture shows no net rotation of plane-polarized light (it is optically inactive), although each of its molecules is chiral.
Students often think Any compound with a C=C bond shows cis–trans isomerism, because rotation about the double bond is restricted. In fact No. Each carbon atom of the C=C bond must carry two different groups. But-1-ene (CH₂= end) and 2,3-dimethylbut-2-ene (two CH₃ groups on each carbon) have no cis–trans isomers.
Students often think If two groups can be drawn on the same side or on opposite sides of a carbon chain, the molecule has cis and trans isomers. In fact No. Rotation about C–C single bonds is free, so the chlorine atoms can take any relative position; the different arrangements are the same compound.
S3.2.8 Fragmentation in mass spectrometry HL
Fragmentation in mass spectrometry
In a mass spectrometer, an organic molecule is ionized by losing an electron to form the molecular ion, M⁺, which can break into a positively charged fragment ion and an uncharged fragment. Only charged species are detected. The m/z values of the fragment ions, and the masses lost from M⁺, give information about the structural units in the molecule.
Molecular ion (M⁺)
The ion formed when a whole molecule loses one electron. For an ion of charge 1+, its m/z value equals the relative molecular mass of the compound. It is usually the peak at highest m/z apart from small isotope peaks (for example an M+1 peak from ¹³C), and it need not be the tallest peak.
Base peak
The tallest peak in a mass spectrum, from the most abundant ion, whose abundance is set to 100% for comparison. It is often a fragment ion rather than the molecular ion.
Students often think The molecular ion gives the tallest peak, because most molecules pass through the spectrometer without breaking up. In fact No. The tallest peak (base peak) is the most abundant ion, often a stable fragment. The molecular ion is the ion of the whole molecule, usually at the highest m/z apart from small isotope peaks, and it can be small.
Students often think The molecular ion forms when a molecule gains an electron from the electron beam, giving a negative ion, M⁻. In fact By the loss of one electron from the molecule, giving a positive ion, M⁺.
S3.2.9 Infrared (IR) spectroscopy HL
Infrared (IR) spectroscopy
A technique in which a sample absorbs infrared radiation at frequencies that match the vibrations (stretching and bending) of its bonds. A spectrum plots transmittance against wavenumber (the reciprocal of wavelength, in cm⁻¹). Absorptions at characteristic wavenumbers identify types of bond, and so functional groups, present in a molecule.
Characteristic IR absorptions
Bonds absorb in characteristic wavenumber ranges, which are used to identify functional groups, for example: O–H in alcohols 3200–3600 cm⁻¹ (broad, hydrogen-bonded); O–H in carboxylic acids 2500–3000 cm⁻¹ (very broad); C–H 2850–3090 cm⁻¹; C=O 1700–1750 cm⁻¹. The shape of a band matters as well as its position: the broad acid O–H band overlaps the sharper C–H absorptions.
IR absorption by greenhouse gases
A molecule absorbs infrared radiation only if the bond vibration changes its dipole moment. Carbon dioxide, water and methane have such vibrations (carbon dioxide is non-polar overall, but its asymmetric stretch and bending vibrations change its dipole moment), so they absorb infrared radiation emitted by the Earth's surface and act as greenhouse gases. Homonuclear diatomic molecules such as N₂ and O₂ have no vibration that changes a dipole moment and do not absorb it.
Students often think Any broad O–H absorption shows that the compound is an alcohol containing a hydroxyl group. In fact No. Alcohol O–H absorbs at 3200–3600 cm⁻¹, while carboxylic acid O–H gives a very broad band at 2500–3000 cm⁻¹. The position of the band tells them apart.
Students often think Any absorption in the region around 2500–3100 cm⁻¹ can be assigned to either C–H or acid O–H without regard to its shape, since the ranges overlap. In fact By its shape and width: the acid O–H band is very broad, spreading across 2500–3000 cm⁻¹, while C–H absorptions (2850–3090 cm⁻¹) are sharper and narrower. Almost all organic compounds show C–H absorptions.
S3.2.10 Proton nuclear magnetic resonance spectroscopy (¹H NMR) HL
Proton nuclear magnetic resonance spectroscopy (¹H NMR)
A technique that detects hydrogen nuclei (protons) in a strong magnetic field. Hydrogen atoms in different chemical environments absorb at different frequencies, so the spectrum shows one signal for each distinct hydrogen environment in the molecule.
Chemical environment
The position of a hydrogen atom within a molecule, defined by the atoms around it. Hydrogen atoms related by the symmetry of the molecule, such as the nine hydrogen atoms of the three CH₃ groups in (CH₃)₃COH, are in the same environment and give one signal. Hydrogen atoms on oxygen, as in –OH and –COOH, also give signals.
Chemical shift (δ)
The position of a signal in a ¹H NMR spectrum, measured in parts per million (ppm) relative to the signal of tetramethylsilane (TMS), which is set at δ 0. It depends on the chemical environment of the hydrogen atoms (for example, the proton of a –COOH group appears in the range 9.0–13.0 ppm), not on how many hydrogen atoms produce the signal.
Integration trace
The relative area under each ¹H NMR signal, shown by an integration trace, which is proportional to the number of hydrogen atoms in that environment. It gives the ratio of hydrogen atoms in the different environments, for example 9:1 for (CH₃)₃COH.
Students often think Only hydrogen atoms bonded to carbon give ¹H NMR signals; the H in –OH or –COOH is not detected. In fact Yes. The hydrogen of an –OH or –COOH group gives a signal (for –COOH in the range 9.0–13.0 ppm).
Students often think Each chemical environment gives one signal of the same size, because the spectrum shows environments, not atoms. In fact No. The area under each signal is proportional to the number of hydrogen atoms in that environment, so a CH₃ signal has three times the area of a single-H signal.
S3.2.11 Spin–spin splitting (n + 1 rule) HL
Spin–spin splitting (n + 1 rule)
A ¹H NMR signal is split by the hydrogen atoms on neighbouring (adjacent) carbon atoms. A signal from hydrogen atoms with n equivalent neighbouring hydrogen atoms is split into n + 1 peaks: n = 0 gives a singlet, 1 a doublet, 2 a triplet and 3 a quartet. Hydrogen atoms in the same environment do not split each other, and the number of hydrogen atoms producing the signal does not affect its splitting. In CH₃CH₂– groups the CH₃ signal is a triplet and the CH₂ signal a quartet.
Students often think n is the number of hydrogen atoms in the group itself, so a CH₃ signal is always a quartet and a CH₂ signal a triplet. In fact No. n is the number of hydrogen atoms on the adjacent carbon atoms. A CH₃ group next to a CH₂ group gives a triplet (n = 2).
Students often think n is the number of atoms (or carbon atoms) bonded to the carbon of the group, so a CH₃ group bonded to one carbon gives a doublet. In fact No. n is the number of hydrogen atoms on the neighbouring carbon atoms. A carbon neighbour with no hydrogen atoms, such as a carbonyl carbon, causes no splitting.
S3.2.12 Combined structural analysis HL
Combined structural analysis
Using several sources of data together to determine a structure: the molecular formula (from composition data and M⁺ in the mass spectrum), bonds and functional groups (IR spectrum), structural units (MS fragments), and the number, relative numbers and neighbours of hydrogen environments (chemical shifts, integration and splitting in the ¹H NMR spectrum). A proposed structure must be consistent with every piece of data.
Students often think The number of signals, their splitting and their areas identify a structure on their own; the chemical shift values need not be checked. In fact Yes, by their chemical shifts. For example, the CH₂ quartet of ethyl ethanoate is near δ 4.1 (bonded to the ester oxygen), while the CH₂ quartet of methyl propanoate is near δ 2.3.
Students often think Once one piece of data has identified a functional group, the compound is identified; the remaining data need not be checked against the proposed structure. In fact No. One band or one technique typically limits the compound to a class, not a single structure; the remaining data (other IR bands, MS fragments, NMR signals, areas and splitting) must all be consistent with the structure proposed.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Propan-2-ol has the molecular formula C₃H₈O. Which statement about the formula C₃H₈O is correct?
Answer and reasoning
It is the molecular formula only, since an empirical formula is simpler. — A student who assumes an empirical formula must always be a reduced version of the molecular formula picks this. It is the simplest whole-number ratio, and 3:8:1 cannot be reduced, so here the two formulas are the same.
It is a condensed structural formula, as it lists every atom present. — Listing every atom is not the same as showing structure. C₃H₈O gives no information about which atoms are bonded to which; a condensed structural formula for propan-2-ol is CH₃CH(OH)CH₃.
It identifies propan-2-ol, since no other compound has this formula. — A student who treats a molecular formula as unique to one compound picks this. Propan-1-ol, CH₃CH₂CH₂OH, and methoxyethane, CH₃OCH₂CH₃, are also C₃H₈O: they are structural isomers of propan-2-ol.
It is both the molecular and the empirical formula of propan-2-ol. — The subscripts 3, 8 and 1 have no common factor, so the ratio C:H:O = 3:8:1 is already the simplest whole-number ratio. The empirical formula is therefore identical to the molecular formula.
CH₃COOH — A student who matches the letters COO picks this. In CH₃COOH the singly bonded oxygen carries a hydrogen atom, so this is the carboxyl group of a carboxylic acid (ethanoic acid), not an ester.
CH₃CH₂OCH₃ — This compound has an oxygen joining two carbon atoms but no C=O. That is an alkoxy group, making it an ether (methoxyethane). An ester needs –COO–.
CH₃COOCH₃ — In CH₃COOCH₃ (methyl ethanoate) the carbonyl carbon is bonded to an oxygen atom that is bonded to a carbon atom, the –COO– ester group, R–COO–R′.
CH₃COCH₂CH₃ — This compound has a C=O inside the chain but only one oxygen atom. It is a ketone (butanone) with a carbonyl group, not an ester, which needs –COO–.
3 Methanol, ethanol, propan-1-ol and butan-1-ol are successive members of the homologous series of alcohols. Which statement about these four compounds is correct?
Answer and reasoning
They fit one general formula and have similar chemical properties. — All four fit CₙH₂ₙ₊₁OH and contain the same functional group, the hydroxyl group, so they undergo similar reactions. Their physical properties change gradually along the series.
They share a functional group, so they have the same boiling point. — A shared functional group gives similar chemical properties, not identical physical ones. Their boiling points rise along the series (about 65 °C, 78 °C, 97 °C and 118 °C).
They are structural isomers of each other, as each has one –OH group. — Isomers need the same molecular formula. These compounds differ by CH₂ each time (CH₄O, C₂H₆O, C₃H₈O, C₄H₁₀O), so they are homologues, not isomers.
Their chemical reactions differ widely, as their chain lengths differ. — Chemical properties are set mainly by the functional group, which is the same in all four. Chain length changes physical properties such as boiling point.
4 The boiling points of four successive alcohols are: methanol 65 °C, ethanol 78 °C, propan-1-ol 97 °C and butan-1-ol 118 °C. Which statement explains this trend?
Answer and reasoning
Longer molecules have more H atoms, so they form more hydrogen bonds. — Only the H bonded to O takes part in hydrogen bonding in an alcohol. Each of these molecules has one –OH group; the extra C–H hydrogen atoms do not form hydrogen bonds.
Longer molecules contain more covalent bonds to break on boiling. — Boiling separates whole molecules; no covalent bonds inside the molecules are broken. The trend comes from stronger intermolecular forces.
Heavier molecules need more energy to move, whatever the forces. — Mass rises along the series, but the cause of the higher boiling point is the stronger London forces between larger molecules, which have more electrons.
London forces grow with chain length; each molecule still has one –OH group. — Each molecule has one –OH group, so hydrogen bonding per molecule is similar across the series. Each extra CH₂ adds electrons and contact area, strengthening the London forces between molecules, so more energy is needed to separate them.
3-methylbutanal — A student who numbers from the left-hand end as the formula is written picks this. The aldehyde carbon is always carbon 1, which puts the methyl group on carbon 2.
2-methylbutanol — The suffix -ol means a hydroxyl group (alcohol). This compound has a –CHO group, an aldehyde, which takes the suffix -al.
2-methylbutanal — The longest chain containing the –CHO carbon has four carbon atoms (butanal). The –CHO carbon is carbon 1, so the methyl branch is on carbon 2: 2-methylbutanal.
2-ethylpropanal — A student who traces the chain from the functional-group carbon into the CH₃ branch without checking the other route picks this. The longest chain containing the –CHO carbon runs through the CH₂CH₃ group and has four carbons, so the parent is butanal.
6 A sample of 2-chlorobutane, CH₃CHClCH₂CH₃, is a racemic mixture. Which statement about the sample is correct? HL
Answer and reasoning
It contains equal amounts of the cis and trans isomers of 2-chlorobutane. — 2-Chlorobutane has no C=C bond or ring, so it cannot show cis–trans isomerism. Its chiral carbon gives enantiomers, which are non-superimposable mirror images; cis and trans isomers are not mirror images of each other.
Its molecules are achiral, which is why the sample shows no optical activity. — Every molecule in the sample is chiral. The sample is optically inactive because the two enantiomers rotate the plane by equal angles in opposite directions, so their effects cancel.
It contains equal amounts of the two enantiomers of 2-chlorobutane. — A racemic mixture contains equal amounts of the two enantiomers. Carbon 2 of 2-chlorobutane is chiral (bonded to H, Cl, CH₃ and C₂H₅), so the compound exists as a pair of enantiomers, which rotate plane-polarized light equally in opposite directions.
Fractional distillation separates it into its two enantiomers. — Enantiomers have identical boiling points, so distillation cannot separate them. They differ only in the direction in which they rotate plane-polarized light and in chiral environments.
7 Which statement about the molecular ion peak in the mass spectrum of an organic compound is correct? HL
Answer and reasoning
It is the tallest peak, as most molecules reach the detector unbroken. — The tallest peak is the base peak, often a fragment ion. For propanone the base peak is at m/z 43 while M⁺ is at m/z 58.
It is due to an ion formed when a molecule gains an electron. — The molecular ion forms when a high-energy electron knocks an electron out of the molecule, giving a positive ion, M⁺.
Any peak at lower m/z than it shows an impurity in the sample. — Peaks at lower m/z are fragment ions from the compound itself, formed when molecular ions break apart. They are evidence of its structure, not of an impurity.
Its m/z value gives the relative molecular mass of the compound. — The molecular ion, M⁺, is the whole molecule minus one electron. With a charge of 1+, its m/z value equals the relative molecular mass of the compound.
8 Carbon dioxide and water vapour are greenhouse gases, but nitrogen and oxygen are not. Which statement explains this difference? HL
Answer and reasoning
CO₂ and H₂O are polar molecules, but N₂ and O₂ are non-polar molecules. — CO₂ is linear and non-polar overall, yet it is a greenhouse gas. What matters is a change in dipole moment during a vibration, not a permanent dipole.
Vibrations of CO₂ and H₂O change their dipole moment, so they absorb IR. — A molecule absorbs infrared radiation when a vibration changes its dipole moment. CO₂ (asymmetric stretch and bending) and H₂O have such vibrations and absorb infrared radiation emitted by the Earth's surface. N₂ and O₂ have none.
Like the ozone layer, CO₂ and H₂O absorb the Sun's ultraviolet radiation. — The greenhouse effect involves infrared radiation emitted by the Earth's surface. Absorption of ultraviolet radiation by ozone is a separate process.
CO₂ and H₂O reflect infrared radiation back to Earth, but N₂ and O₂ do not. — Greenhouse gases absorb infrared radiation, which makes their bonds vibrate, and then re-emit it in all directions. They do not act as a mirror.
9 Which describes the ¹H NMR spectrum of 2-methylpropan-2-ol, (CH₃)₃COH? HL
Answer and reasoning
one signal, as only H atoms on carbon give signals — The hydrogen of the –OH group also gives a signal, so there are two signals, not one.
two signals, with areas in the ratio 1:1 — The number of signals is right, but the area under each signal is proportional to the number of H atoms in that environment: 9 for the CH₃ groups and 1 for OH, not 1:1.
two signals, whose areas are in the ratio 9:1 — The nine hydrogen atoms of the three equivalent CH₃ groups form one environment, and the O–H hydrogen forms a second. Two signals, with areas in the ratio 9:1.
ten signals, one for each H atom in the molecule — Each chemical environment gives one signal, not each hydrogen atom. The nine CH₃ hydrogen atoms are equivalent and give a single signal.
10 In a high-resolution ¹H NMR spectrum, one signal is split into a quartet. What does this show about the structure? HL
Answer and reasoning
The carbon atom(s) adjacent to the group carry three hydrogen atoms in total. — By the n + 1 rule, four peaks means n = 3 hydrogen atoms on the adjacent carbon atom(s), for example a CH₂ group next to a CH₃ group. The number of hydrogen atoms in the group itself is shown by the area, not the splitting.
The group giving the signal is a CH₃ group, since its three H atoms give 3 + 1 = 4 peaks. — n in the n + 1 rule counts hydrogen atoms on the neighbouring carbon atoms, not in the group itself. A CH₃ group next to a carbon with no hydrogen atoms gives a singlet.
The carbon atom giving the signal is bonded to three other carbon atoms in the chain. — Splitting is caused by hydrogen atoms on adjacent carbon atoms, not by the number of atoms bonded to the carbon of the group. n counts neighbouring H atoms.
The environment giving the signal contains four equivalent hydrogen atoms. — The number of hydrogen atoms in an environment is given by the area under the signal. Four peaks means three hydrogen atoms on the adjacent carbon atom(s), whatever the size of the group itself.
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30 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 The skeletal formula of a compound is a zig-zag line of five straight segments joined end to end. An –OH group is drawn at the point where the first and second segments meet (the second point from the left-hand end). No other atoms are shown. What is the molecular formula of the compound?
Answer and reasoning
C₅H₁₂O — A student who counts each line segment as a carbon atom, giving five carbons, picks this. The segments are C–C bonds; the carbon atoms are at the six points where segments end or meet.
C₆H₁₄O — Five segments have six points (two ends and four bends), and each point is a carbon atom, so the chain has six carbon atoms. Each carbon carries enough H atoms to make four bonds: CH₃ at each end, CH(OH) at point 2 and CH₂ at points 3, 4 and 5, giving 3 + 1 + 2 + 2 + 2 + 3 = 13 H on carbon plus 1 H on oxygen. The compound is hexan-2-ol, C₆H₁₄O.
C₄H₁₀O — A student who counts only the four bends as carbon atoms picks this. The two free line ends are also carbon atoms (CH₃ groups), so the chain has six carbons.
C₆H₁₅O — A student who gives every end carbon three H atoms and every inner carbon two, without adjusting for the carbon that carries the –OH, picks this. That carbon already has a bond to oxygen, so it has only one H atom; the total is 14 H, not 15.
Working Carbon atoms: 5 segments give 5 + 1 = 6 points, each a carbon atom. Hydrogen atoms (4 bonds per carbon): point 1 CH₃ (3), point 2 CH bonded to OH (1), points 3–5 CH₂ (2 + 2 + 2), point 6 CH₃ (3) → 13 H on carbon; plus 1 H in –OH → 14 H. One O atom. Molecular formula C₆H₁₄O (hexan-2-ol).
2 Which statement gives the defining feature of an unsaturated organic compound?
Answer and reasoning
Its molecules contain at least one carbon–carbon double or triple bond. — An unsaturated compound contains at least one C=C or C≡C bond. A saturated compound contains only single bonds between its carbon atoms.
Its molecules have fewer H atoms than the alkane with the same number of C atoms. — A low hydrogen count is not the defining feature. Cyclohexane, C₆H₁₂, and chloroethane, C₂H₅Cl, both have fewer H atoms than the matching alkane, yet contain only C–C single bonds and are saturated.
Some of the carbon atoms in its molecules have formed fewer than four bonds. — Each carbon in a C=C bond forms four bonds: two to the other carbon and two to other atoms. 'Unsaturated' refers to the multiple bond, not to a spare, unused bond.
Its molecules contain a functional group, such as –OH or –Cl, in place of an H atom. — Having a functional group does not make a compound unsaturated. Ethanol and chloroethane have only single bonds between carbon atoms and are saturated.
3 Which pair of compounds are successive members of the same homologous series?
Answer and reasoning
CH₃CH₂CHO and CH₃COCH₂CH₃ — These differ by CH₂ (C₃H₆O and C₄H₈O), but propanal is an aldehyde and butanone is a ketone. Successive members must also share the same functional group.
CH₃COOH and CH₃CH₂COOH — Ethanoic acid and propanoic acid are both carboxylic acids (same functional group, general formula CₙH₂ₙO₂) and their formulas differ by one CH₂ unit, so they are successive members of one series.
CH₃CH₂CH₂Cl and CH₃CHClCH₃ — These have the same molecular formula, C₃H₇Cl. They are structural (position) isomers, not homologues, which differ by CH₂.
CH₃CH₂CH₂CH₃ and CH₃CH=CHCH₃ — Butane and but-2-ene differ by H₂ and belong to different series, the alkanes and the alkenes. A homologous series keeps the functional group and adds CH₂.
1-chloro-2-methylbutane — The longest chain has four carbon atoms (butane). Numbering from the CH₂Cl end gives the locant set {1, 2}, lower than {3, 4} from the other end. Prefixes are written alphabetically: 1-chloro-2-methylbutane.
4-chloro-3-methylbutane — A student who numbers from the left-hand end as the formula is written picks this. Numbering from the other end gives the lower locants, 1 and 2.
1-chloro-2-ethylpropane — A student who traces the chain from the functional-group carbon into the CH₃ branch without checking the other route picks this. The longest chain runs through the CH₂CH₃ group and has four carbons; an ethyl branch on carbon 2 is a sign that a longer chain was missed.
2-methyl-1-chlorobutane — A student who places the halogen next to the parent name, as the -ol suffix is in 2-methylbutan-1-ol, picks this. Chloro- is a prefix like methyl-, and prefixes go in alphabetical order: 1-chloro-2-methylbutane.
but-1-en-4-ol — A student who gives the C=C bond the lowest number picks this. The group named as the suffix (-ol) has priority for the lowest locant; the double bond is numbered after it.
but-3-en-1-ol — The chain has four carbon atoms. The –OH group, named by the suffix -ol, takes the lowest locant, 1. The C=C bond is then between carbons 3 and 4 and takes the lower number, 3: but-3-en-1-ol.
but-4-en-1-ol — The –OH locant is right, but a C=C bond between carbons 3 and 4 takes the lower number, 3.
but-3-en-1-al — The suffix -al denotes an aldehyde (–CHO). This compound has a hydroxyl group, –OH, which takes the suffix -ol.
CH₃CH₂CH(OH)CH₂CH₃ — The –OH is on carbon 3, but the class is not the locant. The carbon carrying –OH is bonded to two other carbons, so pentan-3-ol is a secondary alcohol.
CH₃CH(CH₃)CH₂OH — This molecule has a carbon bonded to three carbons, but it is not the carbon carrying –OH. That carbon, the CH₂, is bonded to one carbon, so 2-methylpropan-1-ol is primary.
HOCH₂CH(OH)CH₂OH — Three –OH groups do not make an alcohol tertiary. The class depends on the carbon bearing each –OH: here two are primary and one is secondary.
(CH₃)₂C(OH)CH₂CH₃ — In 2-methylbutan-2-ol the carbon carrying –OH is bonded to three other carbon atoms (two CH₃ groups and a CH₂CH₃ group), so it is a tertiary alcohol.
7 Which statement about the compound (CH₃)₃CNH₂ is correct?
Answer and reasoning
It is a tertiary amine, as the C bonded to N carries three CH₃ groups. — A student who uses the rule for alcohols and halogenoalkanes picks this. For amines, count the carbon atoms bonded directly to nitrogen: there is only one.
It is a secondary amine, as its N atom is bonded to two H atoms. — The class counts carbon atoms on the nitrogen, not hydrogen atoms. An –NH₂ group has one carbon on N and is primary.
It is a primary amine, as its N atom is bonded to one C atom. — Amines are classified by the number of carbon atoms bonded to the nitrogen atom. Here the nitrogen is bonded to one carbon (and two H atoms), so the compound is a primary amine.
It is an amide, since an –NH₂ group is bonded to its C atom. — An amide needs the nitrogen bonded to a carbonyl carbon (–CONH₂). The carbon bonded to N here has no C=O, so this is an amine.
8 Which compound is a structural isomer of butan-1-ol, CH₃CH₂CH₂CH₂OH?
Answer and reasoning
CH₃CH₂OCH₂CH₃ — Ethoxyethane is C₄H₁₀O, the same molecular formula as butan-1-ol, but its atoms are connected differently (an alkoxy group instead of a hydroxyl group). It is a functional group isomer.
CH₃CH₂CH₂CHO — Butanal has four carbons and one oxygen, but its formula is C₄H₈O, two H atoms fewer than butan-1-ol (C₄H₁₀O). Isomers must have the same molecular formula.
HOCH₂CH₂CH₂CH₃ — This is butan-1-ol written from the other end. The atoms are connected in the same order, so it is the same compound, not an isomer.
CH₃(CH₂)₄OH — Pentan-1-ol is C₅H₁₂O, one CH₂ more than butan-1-ol. It is the next member of the same homologous series, not an isomer.
9 Which compound can exist as cis and trans isomers? HL
Answer and reasoning
2,3-dimethylbut-2-ene — A C=C bond is necessary but not sufficient. Each carbon of this C=C carries two identical CH₃ groups, so swapping sides gives the same molecule.
1,3-dichloropropane — This is an open chain with only single C–C bonds. Free rotation lets the chlorine atoms take any relative position, so there are no cis and trans isomers.
2-bromo-3-methylbutane — Carbon 2 of this molecule is chiral, so it has a pair of enantiomers. That is optical isomerism, not cis–trans isomerism, which needs a C=C bond or a ring.
1,2-dimethylcyclobutane — The ring restricts rotation, and carbons 1 and 2 each carry two different groups (a CH₃ group and an H atom). The two CH₃ groups can be on the same side (cis) or opposite sides (trans) of the ring.
propan-2-ol — Carbon 2 carries the –OH group, but its other groups are H and two identical CH₃ groups. A chiral carbon needs four different groups.
butan-2-ol — Carbon 2 of butan-2-ol, CH₃CH(OH)CH₂CH₃, is bonded to four different groups: H, OH, CH₃ and CH₂CH₃. The molecule is therefore chiral and exists as a pair of enantiomers.
but-2-ene — But-2-ene shows cis–trans isomerism, a different kind of stereoisomerism. None of its carbon atoms is bonded to four different groups.
chloroethane — The molecule has different ends, but no carbon atom carries four different groups (each has at least two H atoms), so it is not chiral.
11 The mass spectrum of pentan-3-one, CH₃CH₂COCH₂CH₃ (M_r = 86), has a molecular ion peak at m/z 86 and a peak at m/z 57. Fragment data: m/z 29 = C₂H₅⁺ or CHO⁺; m/z 57 = C₄H₉⁺ or C₂H₅CO⁺. Which species is responsible for the peak at m/z 57? HL
Answer and reasoning
the ion C₂H₅CO⁺, left when C₂H₅ (mass 29) is lost from the molecular ion — Breaking the bond between the carbonyl carbon and one ethyl group gives C₂H₅CO⁺ (m/z 57) and an uncharged C₂H₅ fragment of mass 29 (86 − 57 = 29). Only the charged fragment is detected.
the ion C₂H₅⁺, left behind when a fragment of mass 57 is lost — The m/z value of a peak is the mass of the detected ion, not the mass lost. A peak at m/z 57 is an ion of mass 57; the piece lost has mass 86 − 57 = 29.
an impurity of M_r 57 in the sample, as pure pentan-3-one gives only M⁺ — Peaks below the molecular ion are fragment ions from pentan-3-one itself. The peak at m/z 57 is C₂H₅CO⁺, formed when C₂H₅ (mass 29) is lost from M⁺ (86 − 57 = 29).
the ion C₄H₉⁺, since the data table lists m/z 57 as C₄H₉⁺ — m/z 57 can be C₄H₉⁺ or C₂H₅CO⁺. Pentan-3-one has no unit of four carbons without the oxygen, so C₄H₉⁺ cannot form by breaking one bond; the ion is C₂H₅CO⁺.
Working Mass lost from M⁺ = 86 − 57 = 29, which corresponds to C₂H₅ (2 × 12 + 5 × 1 = 29). Detected ion: C₂H₅CO⁺ = 2 × 12 + 5 × 1 + 12 + 16 = 57. The alternative C₄H₉⁺ (4 × 12 + 9 = 57) would need four carbon atoms bonded together without the oxygen, which pentan-3-one does not contain.
12 A compound, C₃H₆O₂, has an IR spectrum with a very broad absorption across 2500–3000 cm⁻¹ and a strong absorption at 1715 cm⁻¹. Characteristic ranges: O–H (alcohols) 3200–3600 cm⁻¹; O–H (carboxylic acids) 2500–3000 cm⁻¹; C–H 2850–3090 cm⁻¹; C=O 1700–1750 cm⁻¹; C=C 1620–1680 cm⁻¹. Which interpretation is correct? HL
Answer and reasoning
It is an alcohol: the broad band is the O–H of a hydroxyl group. — Alcohol O–H absorbs at 3200–3600 cm⁻¹. A very broad band at 2500–3000 cm⁻¹ is the O–H of a carboxylic acid.
It is an ester: the 2500–3000 cm⁻¹ band is due only to C–H bonds. — C–H absorptions are sharper and lie at 2850–3090 cm⁻¹. A very broad band spreading down to 2500 cm⁻¹ is the hydrogen-bonded O–H of a carboxylic acid.
It is a carboxylic acid: there is a broad acid O–H band and a C=O band. — A very broad band across 2500–3000 cm⁻¹ is the O–H of a carboxylic acid, and 1715 cm⁻¹ lies in the C=O range, 1700–1750 cm⁻¹. The compound is propanoic acid, CH₃CH₂COOH.
It has a C=C bond, which causes the absorption at 1715 cm⁻¹. — C=C absorbs at 1620–1680 cm⁻¹. An absorption at 1715 cm⁻¹ is in the C=O range, 1700–1750 cm⁻¹.
13 A compound, C₂H₄O₂, is either ethanoic acid, CH₃COOH, or methyl methanoate, HCOOCH₃. Its ¹H NMR spectrum shows two singlets, at δ 2.1 and δ 11.5. Typical chemical shifts: R–COOH 9.0–13.0 ppm; H on a carbon bonded to the singly bonded O of an ester, –C(=O)–O–CH₃, 3.7–4.8 ppm. Which conclusion is correct? HL
Answer and reasoning
Methyl methanoate: its two H environments give two singlets. — Methyl methanoate would also give two singlets, so the number and splitting of signals cannot decide. Its CH₃O signal would be in the 3.7–4.8 ppm range; δ 11.5 points to a COOH proton.
Methyl methanoate: an O–H proton gives no signal, so the acid gives one. — The O–H proton of a carboxylic acid does give a signal, in the range 9.0–13.0 ppm. Ethanoic acid therefore gives two signals, matching the spectrum.
Ethanoic acid: δ 11.5 is the CH₃ group, as it has the most H atoms. — The compound is right but the assignment is not. Chemical shift depends on the environment, not the number of H atoms; the number of H atoms shows in the area. δ 11.5 is the COOH proton.
Ethanoic acid: δ 11.5 is the COOH proton and δ 2.1 is the CH₃ group. — A signal at δ 11.5 lies in the range for R–COOH, 9.0–13.0 ppm, and methyl methanoate has no hydrogen in that range. The CH₃ group of ethanoic acid gives the singlet at δ 2.1.
14 In the ¹H NMR spectrum of butanone, CH₃COCH₂CH₃, which splitting pattern does the signal from the CH₃ group bonded to the carbonyl carbon show? HL
Answer and reasoning
a quartet, as the CH₃ group itself contains three H atoms — In the n + 1 rule, n counts H atoms on the neighbouring carbon atoms, not in the group itself. Here the neighbouring carbon has none.
a singlet, as the adjacent carbon has no H atoms — The only carbon adjacent to this CH₃ group is the carbonyl carbon, which has no hydrogen atoms. With n = 0 neighbouring H atoms, the signal is a singlet (n + 1 = 1).
a doublet, as its carbon is bonded to one other carbon atom — n counts hydrogen atoms on neighbouring carbons, not the number of neighbouring carbon atoms. The carbonyl carbon has no H atoms, so there is no splitting.
a triplet, as it is split by the two H atoms of the CH₂ group — The CH₂ group is on the other side of the carbonyl carbon, not on an adjacent carbon. Only H atoms on adjacent carbons cause splitting; the CH₂ splits the other CH₃ signal into a triplet.
15 Compound X, C₄H₈O₂ (M_r = 88), gives these data. IR: sharp absorptions at 2850–3000 cm⁻¹, a strong absorption at 1740 cm⁻¹ and no broad absorption (ranges: C–H 2850–3090 cm⁻¹; C=O 1700–1750 cm⁻¹; O–H in carboxylic acids 2500–3000 cm⁻¹, very broad). MS: M⁺ at m/z 88 and a peak at m/z 29 (CHO⁺ or C₂H₅⁺). ¹H NMR: δ 1.3 (triplet, 3H), δ 2.0 (singlet, 3H), δ 4.1 (quartet, 2H); H on a carbon bonded to the singly bonded O of an ester, –C(=O)–O–CH₂–, appears at 3.7–4.8 ppm. Which identification is consistent with all the data? HL
Answer and reasoning
Ethyl ethanoate, CH₃COOCH₂CH₃: the δ 4.1 quartet is a CH₂ bonded to the ester O and to a CH₃. — IR shows C=O (1740 cm⁻¹) and no O–H. In CH₃COOCH₂CH₃ the O–CH₂ group appears at δ 4.1 (3.7–4.8 ppm) as a quartet (3 H on the adjacent CH₃), the CH₃ of the ethyl group is a triplet, and the CH₃CO group is a singlet; its ethyl group gives C₂H₅⁺ at m/z 29.
Methyl propanoate, CH₃CH₂COOCH₃: its CH₃, CH₂ and OCH₃ give a triplet, quartet and singlet. — The splitting and areas match, but the shifts do not. In CH₃CH₂COOCH₃ the OCH₃ singlet would appear near δ 3.7 and the quartet (CH₂ next to C=O) near δ 2.3; the spectrum has the singlet at δ 2.0 and the quartet at δ 4.1.
Butanoic acid, CH₃CH₂CH₂COOH: the bands at 2850–3000 and 1740 cm⁻¹ are its O–H and C=O. — Sharp bands at 2850–3000 cm⁻¹ are C–H absorptions. A carboxylic acid would give a very broad O–H band across 2500–3000 cm⁻¹ and a ¹H NMR signal at 9.0–13.0 ppm; neither is present.
Propyl methanoate, HCOOCH₂CH₂CH₃: the peak at m/z 29 identifies its HCO group. — m/z 29 could be CHO⁺ or C₂H₅⁺, so it does not decide. HCOOCH₂CH₂CH₃ would give four ¹H NMR signals, not three, and no 3H singlet.
(CH₃)₂CBrCH₂CH₃ — The Br is on carbon 2, but the class is not the locant. The carbon carrying Br is bonded to three other carbons (two CH₃ groups and a CH₂CH₃ group), so 2-bromo-2-methylbutane is tertiary.
CH₃CH₂CHBrCH₂CH₃ — In 3-bromopentane the carbon carrying the Br atom is bonded to two other carbon atoms (two CH₂ groups), so the compound is a secondary halogenoalkane.
CH₃CH₂CH₂CHBr₂ — Two Br atoms do not make a halogenoalkane secondary. The class depends on the carbon carrying the halogen, and here that carbon is bonded to only one other carbon atom, so the compound is not secondary.
CH₃CH₂CH₂CH₂CH₂Br — The class counts carbon atoms, not hydrogen atoms, on the carbon carrying the halogen. That CH₂ carbon is bonded to one other carbon, so 1-bromopentane is primary.
17 The boiling points of the first five straight-chain alkanes are: methane −162 °C, ethane −89 °C, propane −42 °C, butane −1 °C and pentane 36 °C. Which statement about this trend is correct?
Answer and reasoning
Each further CH₂ unit adds the same rise as the last, 37 °C, so hexane is expected to boil at 73 °C. — The steps in the data are 73, 47, 41 and 37 °C, not a constant: each extra CH₂ unit makes a smaller proportional difference to the London forces, so the next rise is less than 37 °C.
The rise is caused by hydrogen bonding, which grows as H atoms are added to the chain. — Alkanes have no O–H or N–H bonds and form no hydrogen bonds. The only forces between alkane molecules are London forces, which strengthen as the number of electrons increases.
The rise per CH₂ unit shrinks along the series, so hexane is expected to boil below 73 °C. — The steps are 73, 47, 41 and 37 °C: each added CH₂ unit strengthens the London forces by a smaller fraction of an already larger molecule, so the next step is smaller than 37 °C and hexane boils below 73 °C (it boils at 69 °C).
The rise occurs because longer molecules have more C–C and C–H bonds to break. — Boiling separates molecules from one another; no covalent bonds are broken. The trend is due to stronger London forces between longer molecules with more electrons.
Working Successive differences: −89 − (−162) = 73 °C; −42 − (−89) = 47 °C; −1 − (−42) = 41 °C; 36 − (−1) = 37 °C. The steps decrease, so the next step is smaller than 37 °C and hexane is expected to boil below 36 + 37 = 73 °C (it boils at 69 °C).
18 Compound Y, C₄H₈O (M_r = 72), gives these data. IR: a strong absorption at 1715 cm⁻¹, sharp absorptions at 2850–3000 cm⁻¹ and no absorption between 3100 and 3700 cm⁻¹ (ranges: C=O 1700–1750 cm⁻¹; C–H 2850–3090 cm⁻¹; O–H in alcohols 3200–3600 cm⁻¹; C=C 1620–1680 cm⁻¹). MS: M⁺ at m/z 72, peaks at m/z 57 and m/z 43. ¹H NMR: δ 1.0 (triplet, 3H), δ 2.1 (singlet, 3H), δ 2.4 (quartet, 2H). Which structure is consistent with all the data? HL
Answer and reasoning
Butanal, CH₃CH₂CH₂CHO: the strong band at 1715 cm⁻¹ with no O–H band identifies the aldehyde C=O group. — The IR band fits an aldehyde as well as a ketone, but the ¹H NMR spectrum does not: butanal has four hydrogen environments (CHO, CH₂, CH₂, CH₃) and no 3H singlet, whereas the spectrum has three signals including a singlet.
2-Methylpropanal, (CH₃)₂CHCHO: its three hydrogen environments give the three signals observed. — The number of signals is right, but the areas and splitting are not. Its two equivalent CH₃ groups would give a 6H doublet and the CH and CHO signals 1H each, not areas of 3:3:2 with a 3H singlet.
Butanone, CH₃COCH₂CH₃: the δ 2.1 singlet is CH₃CO and the ethyl group gives the triplet and quartet. — The C=O band with no O–H fits a ketone or aldehyde. Three signals of areas 3:3:2 with a singlet, a triplet and a quartet fit CH₃CO–CH₂CH₃: the CH₃ next to C=O has no neighbouring H (singlet), and the ethyl group gives a triplet and quartet. Loss of CH₃ (15) gives m/z 57 and loss of C₂H₅ (29) gives m/z 43.
But-3-en-1-ol, CH₂=CHCH₂CH₂OH: the strong absorption at 1715 cm⁻¹ is due to its C=C bond. — C=C absorbs at 1620–1680 cm⁻¹; 1715 cm⁻¹ is in the C=O range. An alcohol would also give a broad O–H band at 3200–3600 cm⁻¹, and none is present.
Working M_r = 4 × 12 + 8 × 1 + 16 = 72, matching M⁺ at m/z 72. Loss of CH₃ (15) gives 72 − 15 = 57, the C₂H₅CO⁺ ion; loss of C₂H₅ (29) gives 72 − 29 = 43, the CH₃CO⁺ ion. Signal areas 3 + 3 + 2 = 8 hydrogen atoms, matching C₄H₈O.
19 The diagram shows the skeletal formula of a hydrocarbon. Which condensed structural formula represents the same compound?
Answer and reasoning
CH₃CH(CH₃)CH₂CH₂CH₃ — You counted the six line segments as six carbon atoms. The lines are bonds; the carbon atoms sit at the line ends and vertices, so the main chain has six carbons, not five, and the branch adds a seventh.
CH₃CH₂(CH₃)CH₂CH₂CH₂CH₃ — You gave the second carbon two H atoms as if it were an ordinary chain carbon. It is bonded to three carbon atoms, so it can carry only one H atom; written as CH₂ with a CH₃ attached it would have five bonds.
CH₃CH(CH₃)CH₂CH₂CH₂CH₃ — You counted every line end and vertex as a carbon atom: six in the main chain and one at the end of the branch on the second carbon. That carbon is bonded to three carbon atoms, so it is CH; the rest are CH₃ at the ends and CH₂ in the chain. The compound is 2-methylhexane, C₇H₁₆.
CH₃CH₂CH₂CH₂CH₂CH₃ — You ignored the short vertical line. Its end is a carbon atom too, so it is a CH₃ branch on the second carbon, and that carbon is CH, not CH₂. The compound has seven carbon atoms, not six.
20 The diagram shows the skeletal formula of a compound. Which statement about its functional groups is correct?
Answer and reasoning
It contains an amido group, since –NH₂ is bonded to a carbon atom. — An amido group is –CONH₂, with the nitrogen bonded directly to the carbonyl carbon. Here the NH₂ is on a CH₂ carbon two bonds away from the C=O, so it is an amino group.
It contains an ester group, since the C=O carbon is bonded to two carbons. — An ester group is –COO–, with the carbonyl carbon bonded to a second oxygen atom. A C=O carbon bonded to two carbon atoms is a ketone carbonyl group.
It is unsaturated, since an –NH₂ group has replaced a hydrogen atom. — Saturated and unsaturated refer to carbon–carbon bonds. All the carbon–carbon bonds here are single, so the compound is saturated whatever functional groups it carries.
It contains amino and ketone groups, since –NH₂ and C=O are on different carbons. — The –NH₂ group bonded to a carbon of the chain is an amino group. The C=O carbon is bonded to two carbon atoms, so it is a ketone carbonyl group. The two groups are on different carbon atoms, so there is no amide.
21 The diagram shows the skeletal formulas of four compounds, P, Q, R and S. Which two are successive members of the same homologous series?
Answer and reasoning
P and Q — Their molecular formulas, C₃H₆O and C₄H₈O, differ by CH₂, but P has an aldehyde group (C=O at the end of the chain) and Q a ketone group (C=O inside the chain). Members of one series share the same functional group.
Q and R — Butanone (Q) and butanal (R) have the same molecular formula, C₄H₈O, with different functional groups: they are functional group isomers, not members of one series.
R and S — Butanal (R, C₄H₈O) and butan-1-ol (S, C₄H₁₀O) have the same number of carbon atoms and differ by H₂; they belong to different series (aldehydes and alcohols). Successive members differ by CH₂ and share a functional group.
P and R — P is propanal, CH₃CH₂CHO, and R is butanal, CH₃CH₂CH₂CHO: both have the aldehyde group –CHO and their formulas differ by one CH₂ unit, so they are successive aldehydes.
22 The diagram shows the skeletal formula of a halogenoalkane. What is its IUPAC name?
Answer and reasoning
3-methyl-2-bromohexane — Halogen and alkyl prefixes are listed together in alphabetical order, so bromo comes before methyl, whatever the locants: 2-bromo-3-methylhexane.
2-bromo-3-methylhexane — Carbon atoms sit at every end and vertex: the longest chain has six, so the parent is hexane. Numbering from the right-hand end gives the substituents the lowest locants, 2 and 3 (rather than 4 and 5 from the left). Prefixes are listed alphabetically: bromo before methyl.
2-bromo-3-propylbutane — The parent chain is the longest continuous chain, which runs through the propyl group: six carbon atoms. A four-carbon chain with a propyl branch is not the longest choice.
5-bromo-4-methylhexane — The chain is numbered from the end that gives the substituents the lowest locants, not from the left-hand end of the drawing. Numbering from the right gives 2 and 3, lower than 4 and 5, so the name is 2-bromo-3-methylhexane.
Working Longest continuous chain: six carbon atoms, so the parent is hexane. Numbering from the right-hand end puts Br on C2 and CH₃ on C3 (locants 2,3); from the left they would be 5,4. Prefixes in alphabetical order: bromo before methyl. Name: 2-bromo-3-methylhexane.
23 The diagram shows the skeletal formula of a compound. What is its IUPAC name?
Answer and reasoning
2-methylpent-2-en-4-ol — The functional group named by the suffix (-ol) takes priority for the lowest locant, so numbering starts from the OH end: 2-ol, 3-ene and 4-methyl.
4-methylpent-4-en-2-ol — A double bond is given the lower of the numbers of its two carbon atoms. Numbered from the OH end, the C=C joins carbons 3 and 4, so it is 3-ene; the methyl group is on carbon 4.
4-methylpent-3-en-2-ol — The hydroxyl group takes the suffix -ol and the lowest locant, so numbering starts at the right-hand end: OH on carbon 2, the C=C between carbons 3 and 4, named by its lower number, 3, and the methyl group on carbon 4.
4-methylpent-3-en-2-al — The suffix -al denotes an aldehyde, –CHO. This compound has a hydroxyl group, –OH, on carbon 2, so the suffix is -ol: 4-methylpent-3-en-2-ol.
Working The suffix -ol takes the lowest locant, so number from the right-hand end: OH on C2, C=C between C3 and C4 (locant 3), methyl on C4. Name: 4-methylpent-3-en-2-ol.
24 The diagram shows four skeletal formulas, P, Q, R and S. Which one represents a structural isomer of butan-2-ol, CH₃CH(OH)CH₂CH₃?
Answer and reasoning
compound P — P is butan-2-ol itself, drawn with the chain running the other way: OH on the third carbon from the left is OH on carbon 2 when numbered from the right. A different drawing is not a different compound.
compound S — S is 2-methylpropan-1-ol, CH₃CH(CH₃)CH₂OH, molecular formula C₄H₁₀O, the same as butan-2-ol but with a branched chain: a structural isomer.
compound Q — Q is but-3-en-2-ol, C₄H₈O. It has four carbon atoms and one oxygen, but two fewer hydrogen atoms than C₄H₁₀O, so its molecular formula differs and it is not an isomer.
compound R — R is pentan-2-ol, C₅H₁₂O, the next member of the same homologous series as butan-2-ol. Isomers must have the same molecular formula.
25 The diagram shows two skeletal formulas, P and Q, of pent-2-ene. Which statement about P and Q is correct? HL
Answer and reasoning
P and Q are the same compound, since the chain can rotate freely about the C=C bond. — Rotation about a C=C bond would require the π bond to break, so it does not occur at ordinary temperatures. The two arrangements cannot interconvert and are different compounds.
P and Q are cis–trans isomers, since rotation about the C=C bond is restricted. — The π bond prevents rotation about C=C, and each double-bond carbon carries two different groups (H and CH₃; H and C₂H₅). In P the CH₃ and C₂H₅ groups are on the same side (cis); in Q on opposite sides (trans).
P and Q are enantiomers, since each is the non-superimposable mirror image of the other. — Neither molecule has a chiral carbon atom, and Q is not the mirror image of P: the two differ in whether the groups are on the same or opposite sides of the C=C. They are cis–trans isomers.
P and Q are structural isomers, since the chain is drawn bent in different directions. — Both have the same connectivity, CH₃CH=CHCH₂CH₃, so they are not structural isomers. The difference is a fixed spatial arrangement about the C=C bond, which makes them stereoisomers.
26 The diagram shows stereochemical formulas of two halogenoalkanes, P and Q. Which statement is correct? HL
Answer and reasoning
Both are chiral, since each has a carbon atom bonded to Cl, H and two alkyl groups. — A carbon atom is chiral only if all four groups differ. In Q the two alkyl groups are both CH₃, so the mirror image of Q is superimposable on Q and Q is not chiral.
Both are chiral, since the mirror image of each, with Cl and H swapped, looks different. — A mirror-image drawing that looks different is not necessarily a different molecule. Q's mirror image can be turned over to superimpose exactly on Q, because two of its groups are the same (CH₃). Only a carbon atom bonded to four different groups, as in P, gives a non-superimposable mirror image.
Neither is chiral, since neither has a C=C bond to restrict rotation. — Restricted rotation about C=C gives cis–trans isomerism, a different kind of stereoisomerism. Chirality arises from a carbon atom bonded to four different groups, which P has.
Only P is chiral, since its central carbon atom carries four different groups. — In P the central carbon is bonded to H, Cl, CH₃ and C₂H₅, four different groups, so it is a chiral centre and P exists as a pair of enantiomers. In Q two of the groups are CH₃, so its mirror image is superimposable.
27 The diagram shows the mass spectrum of propan-1-ol, CH₃CH₂CH₂OH. Which statement about the spectrum is correct? HL
Answer and reasoning
The peak at m/z 31 is the molecular ion, since it is the tallest peak. — The molecular ion is the peak at the highest m/z that corresponds to the whole molecule, m/z 60 here, and it is often small because most molecular ions fragment. The tallest peak is the most stable fragment ion.
The peak at m/z 31 is formed by loss of a fragment of mass 31 from the molecular ion. — An m/z value is the mass of the ion detected, not of the piece lost. Loss of 31 from M⁺ would give a peak at m/z 29; the peak at m/z 31 is the CH₂OH⁺ ion, left after C₂H₅ (mass 29) is lost.
An uncharged C₂H₅ fragment gives the peak at m/z 29 as CH₂OH⁺ forms. — Only charged particles are accelerated and detected. When CH₂OH⁺ forms, the C₂H₅ radical is uncharged and gives no peak; the peak at m/z 29 is the C₂H₅⁺ ion formed in a different fragmentation.
The base peak at m/z 31 is CH₂OH⁺, formed when the molecular ion loses C₂H₅. — M_r of propan-1-ol is 60, so the molecular ion is the small peak at m/z 60. Breaking the C–C bond next to the OH-bearing carbon gives the stable ion CH₂OH⁺ (m/z 31) and an uncharged C₂H₅ fragment (60 − 31 = 29).
Working M_r of propan-1-ol = 3 × 12 + 8 × 1 + 16 = 60, so the molecular ion is at m/z 60. CH₂OH⁺ = 12 + 2 + 16 + 1 = 31, so the base peak at m/z 31 is CH₂OH⁺; the mass lost is 60 − 31 = 29 = C₂H₅ (24 + 5).
28 The diagram shows the IR spectrum of a compound with molecular formula C₄H₈O₂. Characteristic ranges: O–H (alcohols) 3200–3600 cm⁻¹, broad; O–H (carboxylic acids) 2500–3000 cm⁻¹, very broad; C–H 2850–3090 cm⁻¹; C=O 1700–1750 cm⁻¹; C=C 1620–1680 cm⁻¹; C–O 1050–1410 cm⁻¹. To which class does the compound belong? HL
Answer and reasoning
an ester, as the band at 1740 cm⁻¹ is C=O and there is no O–H band — The very strong, narrow band at 1740 cm⁻¹ is C=O, and the strong bands near 1240 and 1050 cm⁻¹ are C–O. Nothing appears at 3200–3600 cm⁻¹ and the absorption at 2850–3000 cm⁻¹ is narrow (C–H), not the very broad acid O–H, so the compound is an ester, such as ethyl ethanoate.
a carboxylic acid, as the band at 2850–3000 cm⁻¹ is the carboxyl O–H — The band near 2980 cm⁻¹ is narrow, as C–H absorptions are. The O–H of a carboxylic acid is very broad, spreading across 2500–3000 cm⁻¹, and no such band is present.
an unsaturated compound, as the band at 1740 cm⁻¹ is a C=C absorption — C=C absorbs at 1620–1680 cm⁻¹ and is usually weak. A very strong band at 1740 cm⁻¹ lies in the C=O range, 1700–1750 cm⁻¹.
an ether, as the strong band at 1240 cm⁻¹ shows a C–O bond — A C–O band is present in ethers and esters alike, but an ether has no C=O bond. The very strong band at 1740 cm⁻¹ shows a C=O group, so the compound is an ester, not an ether.
Working The band at 1740 cm⁻¹ lies in the C=O range, 1700–1750 cm⁻¹. The narrow band near 2980 cm⁻¹ lies in the C–H range, 2850–3090 cm⁻¹; there is no band at 3200–3600 cm⁻¹ and no very broad band across 2500–3000 cm⁻¹, so there is no O–H. Bands at 1240 and 1050 cm⁻¹ lie in the C–O range, 1050–1410 cm⁻¹. C=O plus C–O with no O–H: an ester.
29 The diagram shows the low-resolution ¹H NMR spectrum of propan-1-ol, CH₃CH₂CH₂OH, with the relative area of each signal. Typical chemical shifts: H on a carbon bonded to O, 3.3–4.5 ppm; R–OH, 1.0–6.0 ppm; R–CH₃, 0.9–1.0 ppm. Which statement is correct? HL
Answer and reasoning
The signal at δ 3.6, of area 2, is the CH₂ group bonded to the oxygen atom. — Hydrogen atoms on a carbon bonded to oxygen appear at 3.3–4.5 ppm, and the –CH₂OH group has two hydrogen atoms, matching the 2H signal at δ 3.6.
The signal at δ 2.3, of area 1, cannot be the O–H hydrogen, since O–H hydrogens give no signal. — Every hydrogen atom gives a signal, including the one bonded to oxygen. The single O–H hydrogen gives the 1H signal at δ 2.3, within the R–OH range.
The four signals should have equal areas, since each stands for one chemical environment. — The area under a signal is proportional to the number of hydrogen atoms in that environment: 3 (CH₃), 2 (CH₂), 2 (CH₂O) and 1 (OH), which is what the spectrum shows.
The signal at δ 3.6 is the CH₃ group, since the group with most H atoms has the largest shift. — Chemical shift depends on the environment, not on the number of hydrogen atoms. The signal at δ 3.6 has area 2, and the 3H signal is at δ 0.9, in the R–CH₃ range.
Working Relative areas 3 : 2 : 2 : 1 match CH₃ (3H), CH₂ (2H), CH₂O (2H) and OH (1H). The signal at δ 3.6 lies in the 3.3–4.5 ppm range for H on a carbon bonded to O and has area 2, so it is the –CH₂OH group.
30 The diagram shows the high-resolution ¹H NMR spectrum of chloroethane, CH₃CH₂Cl. Which statement about the splitting is correct? HL
Answer and reasoning
The signal at δ 3.6 is the CH₃ group, because three H atoms give 3 + 1 = 4 lines. — n counts hydrogen atoms on the neighbouring carbon, not in the group itself. The CH₃ group is next to CH₂ (n = 2) and gives the triplet at δ 1.5; the quartet is the CH₂ group.
The signal at δ 3.6 is a quartet because the adjacent CH₃ group carries three H atoms. — The CH₂Cl hydrogens are deshielded by chlorine and appear at δ 3.6. Their only neighbouring carbon is the CH₃ group, with n = 3 hydrogen atoms, so the signal is split into n + 1 = 4 lines.
The signal at δ 1.5 is the CH₂ group, as its carbon has two other atoms bonded to it. — Splitting is not caused by the atoms bonded to the carbon but by hydrogen atoms on adjacent carbons. The CH₂ group is next to CH₃ (n = 3), so it is the quartet, at δ 3.6.
The signal at δ 3.6 shows an environment containing four equivalent H atoms. — The number of lines shows the number of neighbouring hydrogen atoms plus one, not the size of the group. Chloroethane has no group of four equivalent hydrogen atoms; the quartet is CH₂ split by CH₃.
Working CH₂Cl hydrogens: the neighbouring CH₃ carbon carries n = 3 H atoms, so the signal has n + 1 = 4 lines (quartet). CH₃ hydrogens: the neighbouring CH₂ carbon carries n = 2, so 2 + 1 = 3 lines (triplet).
That was your twenty minutes. Real practice on S3.2 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Chemistry guide (first assessment 2025) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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